Trouble computing a sum of Dirichlet characters. Let $\chi(n)$ be a character mod $m$, and let $\rho$ be an $h$th root of unity. I am trying to compute the following sum \begin{equation} \sum_{\chi}(\rho^{-1}\chi(a) + \rho^{-2}\chi(a^2) + \cdots + \rho^{-h}\chi(a^h))\end{equation} and then use the result to prove that the $\phi(m)$ characters mod $m$ at $a$ take all $h$th roots of unity with equal frequency. I know that $\chi(a)$ is an $h$th root of unity, but am not sure how to use this in the computation. Any tips would be appreciated. 
 A: Because $\chi(a)$ is an $h$th root of unity, $\chi(a)^h = 1$, and because $\chi$ is a multiplicative function, $\chi(a)^h = \chi(a^h) = 1$.  We also know that
$$ \sum_\chi \chi(b) =
\begin{cases}
\phi(m) & \text{if $b \equiv 1 \pmod m$, or}\\
0 & \text{otherwise.}
\end{cases}$$
Without loss of generality, we can assume that the order of $a$ is $h$.  Then the sum is
\begin{align}
& \sum_\chi \left(\rho^{-1}\chi(a) + \rho^{-2}\chi(a^2) + \cdots + \rho^{-h}\chi(a^h)\right)\\
& = \rho^{-1} \sum_\chi \chi(a) + \rho^{-2} \sum_\chi \chi(a^2) + \cdots +  \rho^{-h} \sum_\chi \chi(a^h)\\
& = \rho^{-1} \cdot 0 + \rho^{-2} \cdot 0 + \cdots + \rho^{-(h-1)} \cdot 0 + 1 \sum_\chi 1\\
& = \sum_\chi 1\\
& = \phi(m).
\end{align}
That answers your question.  Here is a hint to help you prove that the $\phi(m)$ characters modulo $m$ at $a$ take all $h$th roots of unity with equal frequency:  Find the value of
$$1 + z + z^2 + \cdots + z^{h-1}$$
if $z$ is an $h$th root of unity, and then take $z = \rho^{-1} \chi(a)$. 
